The centre of the ellipse frac{(x+y-3)^2}{9}+ | vedclass

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(D) The given equation of the ellipse is $\frac{(x+y-3)^2}{9}+\frac{(x-y+1)^2}{16}=1$. To find the centre of the ellipse,we set the linear expressions

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$(1,2)$

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(D) The given equation of the ellipse is $\frac{(x+y-3)^2}{9}+\frac{(x-y+1)^2}{16}=1$. To find the centre of the ellipse,we set the linear expressions inside the squares to zero: $x+y-3=0 \quad (i)$ $x-y+1=0 \quad (ii)$ Adding equations $(i)$ and $(ii)$: $(x+y-3) + (x-y+1) = 0 + 0$ $2x - 2 = 0 \Rightarrow x = 1$ Substituting $x=1$ into equation $(i)$: $1+y-3=0 \Rightarrow y=2$ Thus,the centre of the ellipse is $(1,2)$.

The centre of the ellipse $\frac{(x+y-3)^2}{9}+\frac{(x-y+1)^2}{16}=1$ is

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